Question

A solution is prepared by dissolving 0.7236 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the diluted oxalic acid solution?

Answer 1

C H2C2O4 = 3.216 E-3 M

Step-by-step explanation:

m H2C2O4 = 0.7236 g

∴ Mw H2C2O4 = 90.03 g/mol

⇒ n H2C2O4 = (0.7236 g)(mol/90.03 g) = 8.04 E-3 mol

⇒ C H2C2O4 = (8.04 E-3 mol/ 0.1 L ) = 0.0804 mol/L

dilution factor (f):

∴ f = 10 mL/250 mL = 0.04

final molarity:

⇒ C H2C2O4 = (0.0804mol/L)(0.04) = 3.216 E-3 M